TOJ-1339 Steps

最新推荐文章于 2018-10-13 17:51:07 发布

转载最新推荐文章于 2018-10-13 17:51:07 发布 · 172 阅读

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原文链接：http://www.cnblogs.com/shenchuguimo/p/6375498.html

本文探讨了从一整数到另一整数的最短步数算法，步长需为非负且不超过前一步长±1，首尾步长固定为1。通过分析两数之差，利用数学公式快速计算所需步数，适用于编程竞赛和算法优化。

One steps through integer points of the straight line. The length of a step must be nonnegative and can be by one bigger than, equal to, or by one smaller than the length of the previous step.

What is the minimum number of steps in order to get from x to y? The length of the first and the last step must be 1.

Input consists of a line containing n, the number of test cases. For each test case, a line follows with two integers: 0 ≤ x ≤ y < 2³¹. For each test case, print a line giving the minimum number of steps to get from x to y.

Sample Input

Output for Sample Input

3
3
4

Source: Waterloo Local Contest Jan. 29, 2000

在前一个数的基础上依次加若干个数，求最少需要加几个数可得到后一个数。要求加的数的序列中，起始和末尾的数为1，其他的数

比前一个大1，小1，或相等。

规律总结：

　　对所给的两个数的差x，若有自然数i，i(i+1)<x<=(i+1)^2，需乘2i+1个数；若有i^2<x<=i(i+1)，需乘2i个数。

#include <iostream>  
#include <cmath>  
using namespace std;  
  
int main()  
{  
    int x,y,n;  
    cin >> n;  
    while(n--)  
    {  
        cin >> x >> y; 
        int n = y-x;
        int q = (int)(sqrt(y-x)-1e-9);
        if(n<=(q+1)*q) cout<<q*2<<endl;
        else cout<<q*2+1<<endl;
    }  
  
    return 0;  
}

转载于:https://www.cnblogs.com/shenchuguimo/p/6375498.html